Afast food restaurant executive wishes to know how many fast food meals teenagers eat each week. they want to construct a 85% confidence interval for the mean and are assuming that the population standard deviation for the number of fast food meals consumed each week is 1.4 . the study found that for a sample of 502 teenagers the mean number of fast food meals consumed per week is 3.1 . construct the desired confidence interval. round your answers to one decimal place. lower end point and upper end point

The 85% confidence interval for the mean number of fast food meals each week is (3.01 meals, 3.19 meals).

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

In which is the standard deviation of the population and n is the length of the sample. So

The lower end of the interval is the mean subtracted by M. So it is 3.1 - 0.09 = 3.01 meals.

The upper end of the interval is the mean added to M. So it is 3.1 + 0.09 = 3.19 meals.

The 85% confidence interval for the mean number of fast food meals each week is (3.01 meals, 3.19 meals).

The answer is $5,000.73.
Explanation: If one Canadian dollar bill is $1.00 and a Trinidad’s $1.00 is $2.73 then that means if you wanna convert it you have to double the amount including the €73 and you have 2,500 you need to double the amount then add the remaining €73.